How many moles of al(oh)^3 is in 12.4 g of aluminum hydroxide?
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Ответ:
0.1589 mol ≅ 0.16 mol.
Explanation:
Knowing that the no. of moles can be calculated using the relation:no. of moles (n) = mass/molar mass
mass of Al(OH)₃ = 12.4 g & molar mass of Al(OH)₃ = 78.01 g/mol.
∴ n = mass/molar mass = (12.4 g)/(78.01 g/mol) = 0.1589 mol ≅ 0.16 mol.
Ответ:
2NaCl I am sure of it
Explanation:
Synthesis reaction: 2Na + Cl2 --> 2NaCl